Problem
K The annual earnings (in dollars) of 35 randomly selected microbiologists are shown in the data table. Use the data to (a) find the sample mean, (b) find the sample standard deviation, and (c) construct a $98 \%$ confidence interval for the population mean. \begin{tabular}{llllllll} 100,645 & 81,342 & 77,435 & 68,134 & 51,463 & 67,126 & 94,373 & 66,103 \\ 79,349 & 74,452 & 44,450 & 86,482 & 60,708 & 57,421 & 55,282 & 78,497 \\ 47,868 & 98,898 & 80,664 & 93,459 & 63,272 & 75,144 & 51,123 & 60,765 \\ 92,495 & 83,470 & 80,442 & 64,446 & 74,293 & 56,737 & 47,006 & 88,894 \\ 75,481 & 62,041 & 82,126 & & & & & \end{tabular} (a) Find the sample mean. $\bar{x}=\square$ (Type an integer or decimal rounded to two decimal places as needed.) (b) Find the sample standard deviation. $s=\square$ (Type an integer or decimal rounded to two decimal places as needed.) (c) Construct a $98 \%$ confidence interval for the population mean. \[ \square<\mu<\square \] (Round to the nearest integer as needed.)
Solution
Step 1 ：Sample mean: \( \bar{x} \approx 75,103.06 \)
Step 2 ：Sample standard deviation: \( s \approx 14,825.97 \)
Step 3 ：98% confidence interval for the population mean: \( 74,014 < \mu < 76,192 \)
